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The coefficient of x^6   in the expansion of (1+x^2-x^3)^8  is

Option: 1

80 


Option: 2

84 


Option: 3

88 


Option: 4

92 


Answers (1)

best_answer

As we have learned

Coefficient of x^{R} -

We write general term T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}

and a= f\left ( x \right )

\dpi{120} T_{r+1}=^{n} c_{r}\cdot x^{n-r}\left ( f\left ( x \right ) \right )^{r}

We arrange all of x together and make x^{\left ( n,\,r \right )}

compare : x^{\left ( n,c,r \right )}= x^{R}

find r

- wherein

Take a in terms of x.

r can't be negative or fraction.

 

 (1+x^2-x^3 )^8 = (1+x^2)^8 + ^8 C_1 (1+x^2)^7 (-x^3)^1+ ^8 C_2 (1+x^2)^6 (-x^3)^2+....

 

\Rightarrow coefficient\: \: of \: \: x^6 = ^8C_3 -0 + ^8C_2 \cdot ^6C_0 = 84

 

 

 

 

Posted by

Divya Prakash Singh

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